Using the work output of 1.02 x 10^14 J found for that same coal-fired power station, and the 2.50 x 10^14 J of heat transfer into it, what is the station's efficiency, expressed as a percentage?
A14.5%
B24.5%
C59.2%
D40.8%
Answer & Solution
Correct answer: D. 40.8%
1. Efficiency is defined as Eff = W / Qh, the work output divided by the heat transfer into the engine.
2. Substitute the values with units: Eff = (1.02 x 10^14 J) / (2.50 x 10^14 J).
3. The powers of ten cancel, leaving 1.02 / 2.50 = 0.408.
4. Converting to a percentage: 0.408 x 100 = 40.8%.
5. Option C, 59.2%, is the fraction of heat transfer lost to the environment rather than the fraction converted to work; the two percentages must add to 100%.
6. Options A and B do not correspond to dividing 1.02 by 2.50 in the correct order, so neither matches the required substitution.
_Source: OpenStax College Physics (CC BY 4.0), Ch 15 "Thermodynamics", section 15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency_
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